A) 32
B) 33
C) 34
D) 35
Correct Answer: B
Solution :
\[{{T}_{r+1}}={}^{256}{{C}_{r}}{{(\sqrt{3})}^{256-r}}{{(\sqrt[8]{5})}^{r}}\] \[={}^{256}{{C}_{r}}{{(3)}^{\frac{256-r}{2}}}{{(5)}^{r/8}}\] Terms would be integral if \[\frac{256-r}{2}\] and \[\frac{r}{8}\] both are positive integer. As 0 £ r £ 256, \[\therefore r=0,\,8,\,16,\,24,.....,256\] For above values of r, \[\left( \frac{256-r}{2} \right)\] is also an integer.
You need to login to perform this action.
You will be redirected in
3 sec
